Two forces {vec F_1} = 5hat i + 10hat j - 20h | vedclass

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Question

(b) $\cos 	heta = \frac{{\overrightarrow {{F_1}} .\overrightarrow {{F_2}} }}{{|{F_1}||{F_2}|}}$

$ = \frac{{(5\hat i + 10\hat j - 20\hat k).(10\hat

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(b) $\cos 	heta = \frac{{\overrightarrow {{F_1}} .\overrightarrow {{F_2}} }}{{|{F_1}||{F_2}|}}$

$ = \frac{{(5\hat i + 10\hat j - 20\hat k).(10\hat i - 5\hat j - 15\hat k)}}{{\sqrt {25 + 100 + 400} \sqrt {100 + 25 + 225} }}$$ = \frac{{50 - 50 + 300}}{{\sqrt {525} \sqrt {350} }}$

$&rArr;$ $\cos 	heta = \frac{1}{{\sqrt 2 }}$

$	herefore $ $	heta = 45^\circ $

Two forces ${\vec F_1} = 5\hat i + 10\hat j - 20\hat k$ and ${\vec F_2} = 10\hat i - 5\hat j - 15\hat k$ act on a single point. The angle between ${\vec F_1}$ and ${\vec F_2}$ is nearly ....... $^o$

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